\(\left|\begin{array}{cc}\mathrm{x}+1 & \mathrm{x}-1 \\ \mathrm{x}^2+\mathrm{x}+1 & \mathrm{x}^2-\mathrm{x}+1\en…
Q1 FREE PREVIEW
\(\left|\begin{array}{cc}\mathrm{x}+1 & \mathrm{x}-1 \\ \mathrm{x}^2+\mathrm{x}+1 & \mathrm{x}^2-\mathrm{x}+1\end{array}\right|\) is equal to :
✓ Correct answer: b)
2
Explanation
Given: \(\left|\begin{matrix}x+1 & x-1 \\ {x}^{2}+x+1 & {x}^{2}-x+1\end{matrix}\right|\)
\(=\left(x+1\right)\left({x}^{2}-x+1\right)-\left(x-1\right)\left({x}^{2}+x+1\right)\)
\(=x^3-x^2+x+x^2-x+1-(x^3+x^2+x-x^2-x-1)\)
\(=\left(x^3+1\right)-\left(x^3-1\right)=2\)
Practice more Board Maths PYQs
See every question on Determinants, or browse the full Board question bank.
See all questions on Determinants →